The 5DM2 pixel size is approximate 0.00645mm in size and so some people talk about using 0.007mm as the CoC. I think canon said once they use 0.035mm for DOF charts, full frame normal is 0.033mm, Zeiss has the formula of dividing the sensor diagonal size by 1730 which gives 0.025mm, and some people say they use 0.02mm when they are going to a large print.

I understand how magnification and viewing distance effect the CoC, but I've never seen any concrete formulas about how it's done assuming normal eyesight. So why would one advise using the approximate pixel size (0.007mm) for the CoC on a 5DM2? Is this for a specific magnification/viewing distance combinations, and if so how would you go about figuring those out?

In the second link it's mentioned that a 5DM2 8"x10" print is an 8.5x magnification and viewing pixels at 100% on a computer screen is 45x magnification, is the 45x number accurate? How would did they come up with that?

1 Answer
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Most depth of field (DoF) calculations are based on the assumption that the image will be viewed as an 8X10 print at a viewing distance of about 10 inches (25cm) by a person with 20/20 vision. For a 35mm film sized image, that means about an 8X magnification factor. For a blur circle to be perceived as a point at that display size and viewing distance, it must be about .03mm or smaller on the unmagnified virtual image projected by the lens onto the recording medium (the film negative or the digital sensor). Zeiss assumed that some people viewing the photo would have better than 20/20 vision and allowed for that in their calculations to arrive at .025mm. In either case, the allowable CoC for viewing an 8X10 print at 10 inches is several pixel widths wide. In the case of the Canon 5D mark II, both .03 and .025 are between 4 and 5 pixels wide.

If you are pixel peeping at 100%, then you are viewing the image at a much greater magnification than the 8X10 standard. You are viewing it at around 45X magnification, so areas of the image that appear sharp at 8X magnification are revealed to be slightly blurry at 45X magnification. In this case, you would need to use the pixel pitch of the sensor for your circle of confusion when computing what the DoF would be for that viewing condition.

With digital sensors, the size of the pixel determines the size at which the circle of confusion (CoC) becomes significant when viewing at 100% crops. Any blur circle smaller than the pixel pitch will be recorded as a single pixel. Only when the blur circle becomes larger than an individual pixel will it be recorded by two adjacent pixels.

The other thing to consider is that each pixel on your sensor is only sensitive to either Red, Green, or Blue. To produce a color image either your camera's jpeg engine or your RAW conversion software applies a demosaicing algorithm to produce a seperate R, G, & B value for each pixel using complex mathematical interpolation. This demosaiced image can then be sharpened by software that uses contrast between adjacent pixels to try and reclaim some of that lost resolution. That is why .007mm is probably close enough to the .00639 pixel pitch of the 5DII's sensor to use in calculating the DoF when viewing at 100%.

Here is another way to look at it. If you have a monitor with a resolution of 1920x1080 and you display an uncropped image from your 5D II on it, each pixel of you monitor is combining between 3 and 4 pixels worth of data from your camera into each pixel. Any part of the original image that is blurred by less than 3 pixels will appear just as sharp on your computer monitor as the sharpest part of the picture. But when you magnify the picture to 100% and only look at part of the image filling your entire screen, you will be able to see any blur that is greater than 1 pixel wide (assuming your vision is good enough and you are close enough to see individual pixels).

Amazing answer I love it, can you explain how you reached the 45x magnification number?
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DavyCrockettJun 2 '13 at 2:48

This was also helpful; rags-int-inc.com/PhotoTechStuff/DoF but it says "For digital sensors, the CoC cannot be smaller than the physical size of two pixels (image elements). Obviously nothing smaller can be resolved. Typical pixel sizes for high resolution digital cameras are in the range of .006 to .012mm... These equate to CoC values of .012 and .023mm." Is he correct in saying you should not use less than two pixels in size? It seems like 1 pixel is the min. detail the sensor could resolve, and detail larger would be split to a second pixel?
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DavyCrockettJun 2 '13 at 3:04

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@DavyCrockett - the "45x" depends on the monitor resolution. That particular figure assumes something like a 72-75 pixel-per-inch monitor; 100 ppi would yield the equivalent of about 38x (depending on the linear resolution of the camera -- this is specifically for a 20-ish MP full-frame (24x36mm) sensor). You'd need a 72"-class 16:9 monitor at 100-ish ppi (4K) to see the whole image at 100%, equivalent to print of between 36"x54" and 40"x60". That's just a little bit bigger than the 8x10 assumption for the old calculations.
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user2719Jun 2 '13 at 14:12

@DavyCrockett I think using two pixels makes sense, by analogy with the diffraction limit - the point where you can no longer tell two airy disks apart doesn't happen until the airy disk diameter reaches two pixels. Similarly, a CoC of more than 1 pixel will bleed over and reduce the contrast between a pixel and its neighbour, but actual Confusion, the point where you can't tell two pixels apart, doesn't happen until CoC reaches two pixels. That would be my best guess, anyway.
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j-g-faustusJun 2 '13 at 14:32

But if a blur circle one pixel wide isn't exactly centered over one pixel, it is already bleeding onto between one and three additional adjacent pixels. Likewise, if a blur circle two pixels wide isn't centered on the junction of a 4X4 set of pixels it could be bleeding over as many as nine pixels (3X3).
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Michael ClarkJun 2 '13 at 17:37